Untitled Goose Game

Let us see if this is true
Proof.
No

Lemma 1.

Let . Then

Proof.


The bigger problem is the claim that the area of the sector is . This is generally shown by saying that the area of the circle is , so an angle of radian cuts off a sector of area Since the radius is 1 here, the result follows.

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The next image is also supposed to be svg, but is tikz :(

OPAXYsin(x)tan(x)1xl

Problems with the above proof
...

There are a couple of small and a large problem with the above proof that are generally not addressed in most books.

  1. The fact that the area of is less than or equal to the area of the sector , while clear from the picture, requires a formal proof. Given that 'area' satisfies , it suffices to show that the triangle is a subset of the sector (why area satisfies that property comes from the area being the Lebesgue measure on ). That is very easily done.

  2. Similarly although a bit less of an issue is that no argument is generally given for showing that the area of the sector is less than or equal to the area of and is just asked to refer to the picture. We shall give a proof for this as well.

  3. The continuity of is also a small issue that is generally taken for granted. While intuitively clear, one does need a proof for it. It easily follows from the setup and we will see how one goes about doing it.

  4. The bigger problem is the claim that the area of the sector is . This is generally shown by saying that the area of the circle is , so an angle of radian cuts off a sector of area Since the radius is 1 here, the result follows.
    But how do we know that the area of the circle is ? Well, the go-to method is to prove it via integrals.

    a. A standard proof goes as follows.

    The circle of radius is given by the equation . So, the area is given by

    But here we have used that the derivative of is whose proof also uses the limit in question. Thus this proof for the area of a circle cannot be used.

    b. There is also another proof that tries to avoid differentiating any trigonometric function. This is the so-called shell method, where we divide the circle into very small rings (infinitesimally small) of radius and thickness . These are of area and integrate over them to get
    But this also hides the derivatives when looking through the lens of rigor. Where are the derivatives? Well, the shell method is rigorously proven by the change-of-variables formula. In this case, we change variables from to (Cartesian to polar). The Jacobian of this transformation is where the derivatives are hidden.

    c. How about a non-integral proof? Like say Archimedes' proof, where he shows that the area of the circle is neither greater than nor less than the area of a triangle with base equal to the circumference and height equal to the radius?

    If one reads very carefully, the proof assumes that if a curve converges pointwise to another curve (in this case, inscribed polygons converging to a circle), then so does the areas. This is not rigorously justified and can fall prey to the staircase paradox. The rigorous justification again uses the limit in question, so... we get the point.

We need to answer each of these questions if we are to make sure the calculation of the limit above is airtight. We proceed to do exactly that.

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Is the above getting rendered? The commutative diagram I mean.

OPAXYsin(x)tan(x)1xl